Based on VSEPR theory, which of the following sequences correctly...

In $\mathrm{H}_2\mathrm{S}$, the central atom sulfur is $sp^3$ hybridized and has two lone pairs. Due to strong lone pair-lone pair repulsions, the bond angle decreases significantly to approximately $92^\circ$.

In $\mathrm{CH}_4$, the central atom carbon is $sp^3$ hybridized with no lone pairs. The geometry is perfectly tetrahedral with a bond angle of $109.5^\circ$.

In $\mathrm{BF}_3$, the central atom boron is $sp^2$ hybridized with no lone pairs. The geometry is trigonal planar with a bond angle of $120^\circ$.

In $\mathrm{BeF}_2$, the central atom beryllium is $sp$ hybridized with no lone pairs. The geometry is linear with a bond angle of $180^\circ$.

Therefore, the increasing order of bond angles is $\mathrm{H}_2\mathrm{S} < \mathrm{CH}_4 < \mathrm{BF}_3 < \mathrm{BeF}_2$.

Answer: $\mathrm{H}_2\mathrm{S} < \mathrm{CH}_4 < \mathrm{BF}_3 < \mathrm{BeF}_2$